Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $a \neq 0$. $k = \dfrac{a + 1}{a - 4} \times \dfrac{a^2 - 2a - 8}{-10a^2 - 10a} $
Explanation: First factor the quadratic. $k = \dfrac{a + 1}{a - 4} \times \dfrac{(a - 4)(a + 2)}{-10a^2 - 10a} $ Then factor out any other terms. $k = \dfrac{a + 1}{a - 4} \times \dfrac{(a - 4)(a + 2)}{-10a(a + 1)} $ Then multiply the two numerators and multiply the two denominators. $k = \dfrac{ (a + 1) \times (a - 4)(a + 2) } { (a - 4) \times -10a(a + 1) } $ $k = \dfrac{ (a + 1)(a - 4)(a + 2)}{ -10a(a - 4)(a + 1)} $ Notice that $(a + 1)$ and $(a - 4)$ appear in both the numerator and denominator so we can cancel them. $k = \dfrac{ \cancel{(a + 1)}(a - 4)(a + 2)}{ -10a\cancel{(a - 4)}(a + 1)} $ We are dividing by $a - 4$ , so $a - 4 \neq 0$ Therefore, $a \neq 4$ $k = \dfrac{ \cancel{(a + 1)}\cancel{(a - 4)}(a + 2)}{ -10a\cancel{(a - 4)}\cancel{(a + 1)}} $ We are dividing by $a + 1$ , so $a + 1 \neq 0$ Therefore, $a \neq -1$ $k = \dfrac{a + 2}{-10a} $ $k = \dfrac{-(a + 2)}{10a} ; \space a \neq 4 ; \space a \neq -1 $